已知向量m=(4cosx,-2sinπ/6),n=(sin(x+π/3),2cosπ/6),函数f(x)=m*n,x∈R求f(x)取最大值时的x值若函数g(x)=f(x+φ)是偶函数,0
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![已知向量m=(4cosx,-2sinπ/6),n=(sin(x+π/3),2cosπ/6),函数f(x)=m*n,x∈R求f(x)取最大值时的x值若函数g(x)=f(x+φ)是偶函数,0](/uploads/image/z/8891200-64-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fm%3D%284cosx%2C-2sin%CF%80%2F6%29%2Cn%3D%28sin%28x%2B%CF%80%2F3%29%2C2cos%CF%80%2F6%29%2C%E5%87%BD%E6%95%B0f%28x%29%3Dm%2An%2Cx%E2%88%88R%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%8F%96%E6%9C%80%E5%A4%A7%E5%80%BC%E6%97%B6%E7%9A%84x%E5%80%BC%E8%8B%A5%E5%87%BD%E6%95%B0g%28x%29%3Df%28x%2B%CF%86%29%E6%98%AF%E5%81%B6%E5%87%BD%E6%95%B0%2C0)
已知向量m=(4cosx,-2sinπ/6),n=(sin(x+π/3),2cosπ/6),函数f(x)=m*n,x∈R求f(x)取最大值时的x值若函数g(x)=f(x+φ)是偶函数,0
已知向量m=(4cosx,-2sinπ/6),n=(sin(x+π/3),2cosπ/6),函数f(x)=m*n,x∈R
求f(x)取最大值时的x值
若函数g(x)=f(x+φ)是偶函数,0<φ<π/2,求g(x)的表达式
已知向量m=(4cosx,-2sinπ/6),n=(sin(x+π/3),2cosπ/6),函数f(x)=m*n,x∈R求f(x)取最大值时的x值若函数g(x)=f(x+φ)是偶函数,0
m*n=4cosxsin(x+π/3)-2sinπ/6*2cosπ/6
=4cosx(2分之1sinx+2分之根号3cosx)-根号3
=2cosxsinx+2倍根号3cosx的平方-根号3(然后利用二倍角公式)
=sin2x+根号3倍cos2x(然后利用合一变形)
=2sin(2x+π/3)
所以当2x+π/3=π/2+2kπ时,f(x)取到最大值
所以x+=π/12+kπ(k取整数)
若函数g(x)=f(x+φ)是偶函数,0<φ<π/2,则f(x+φ)=2cos2x
f(x+φ)=2sin(2(x+φ)+π/3)=2sin(2x+2φ+π/3)(利用诱导公式)
所以2φ+π/3=π/2
所以φ=π/12
f(x)=4cosxsin(x+π/3)-2sinπ/6*2cosπ/6
=4cosx(sinxcosπ/3+sinπ/3cosx)-4*1/2*1/2
=4cosx(1/2sinx+√3/2cosx)-1
=2sinxcosx+2√3cosxcosx-1
=sin2x+√3(1-cos2x)-1
...
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f(x)=4cosxsin(x+π/3)-2sinπ/6*2cosπ/6
=4cosx(sinxcosπ/3+sinπ/3cosx)-4*1/2*1/2
=4cosx(1/2sinx+√3/2cosx)-1
=2sinxcosx+2√3cosxcosx-1
=sin2x+√3(1-cos2x)-1
=sin2x-√3cos2x+√3-1
=2sin(2x-π/3)+√3-1
因为g(x)为偶函数
所以g(x)=g(-x)
sin(2x-π/3+φ)=cos(2x)
所以φ-π/3=-π/2
φ=π/6
(0<φ<π/2)
故φ-π/3=π/2舍去
所以g(x)=2cos2x+√3-1
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